We conclude this section with a proof of the
proposition 8. The group of units in
/pe is of
order
pe - pe - 1 = pe - 1(p - 1), thus it is enough to find elements
of order p - 1 and pe - 1 in this group. First we prove a result
that will be useful in other contexts

using
2kk + 1 since k 1. We are given
f (nk) = ckpk for
some constant ck. Moreover,
nk = n(mod p) so
f'(nk) = f'(n)(mod p) is an invertible element of
/p. Let m be an inverse so that
mf'(nk) = 1(mod p). We
put bk = - mck and obtain the required condition.

We now apply this to the polynomial
f (T) = Tp - 1 - 1 and any integer
n prime to p to conclude that there is an integer ne so that
nep - 1 = 1(mod pe) (note that
f'(T) = - Tp - 2(mod p)). This
gives us the required elements of order p - 1 (since such exist modulo
p). Moreover, we see that the units in
/pe can be written
as gau where g is an element of order (p - 1) and
u = 1(mod pe). Let U1 be the group of elements of the latter
kind. We will now apply log and exp in a suitable way to
conclude the result.

Lemma 11

Let x be divisible by p then the power of p that divides
xn/n is at least
n - [logp(n)], where the latter term denotes
the integral part of logp(n). In particular, this goes to
infinity with n.

Let x be divisible by p, then the power of p that divides
pn/n! is at least
n - [n/pi]. In particular, if p
is odd then the latter term goes to infinity with n.

Proof.
Exercise.

It follows that only finitely many terms of the power series

log(1 - x) = -

survive in
/pe when we
substitute x by a multiple of p. Thus we obtain a map

log : U1/pe

which in fact takes values in the ideal
p/pe, which
isomorphic to the additive group
/pe - 1. Elementary
manipulations of the power series combined with the binomial theorem
and the fact that all but finitely many terms are zero can be used to
show that
log(1 + x . y + x + y) = log(1 + x) + log(1 + y). Similarly,
for p odd we obtain a map

exp : p/peU1

by means of the usual power series

exp(x) =

which satisfies
exp(x + y) = exp(x) . exp(y). We also check by
direct substitution that
log(exp(x)) = x and vice versa. It follows
that the group U1 is isomorphic to the group
p/pe which
is in turn isomorphic to
/pe - 1. We note in passing that
a generator g of U1 of order pe - 1 corresponds to the generator
p in
p/pe via the expression g = exp(p)!